Questions
A thin rod AB of mass 1 Kg and length 2m can rotate about a spindle passing through its mid point C and perpendicular to its length. Now a 2 N-m torque is applied to the rod for 10 seconds. Then angular momentum of the rod at the end of 10 S is
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Correct option is A
dLdt=τ, ∫0LdL=τ∫0tdt⇒L=τt=2×10 N−m−s=20 N−m−sSimilar Questions
The moment of inertia (I) and the angular momentum (L) are related by the expression [J&K CET 2013]
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